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A science-fiction tale describes an artificial "planet" in the form of a band completely encircling a sun (Fig. \(6-31\) ). The inhabitants live on the inside surface (where it is always noon). Imagine that this sun is exactly like our own, that the distance to the band is the same as the Earth-Sun distance (to make the climate temperate), and that the ring rotates quickly enough to produce an apparent gravity of \(g\) as on Earth. What will be the period of revolution, this planet's year, in Earth days?

Short Answer

Expert verified
The artificial planet's year is approximately 365 Earth days.

Step by step solution

01

Understanding the Scenario

We need to determine the period of revolution for an artificial band encircling a sun, equivalent to the Earth's year in days. This band revolves around the sun at the same distance as Earth from the Sun and rotates to create an apparent gravity of \( g \).
02

Using Centripetal Force for Artificial Gravity

For the band to provide gravity \( g \), it must rotate quickly enough that the centripetal acceleration is equal to \( g \). The centripetal acceleration formula is \( a_c = \frac{v^2}{r} \), where \( v \) is the linear velocity of the band and \( r \) is the radius of the orbital path (same as Earth's orbital radius).
03

Relate Gravity to Velocity

Set the centripetal acceleration equal to Earth's gravity: \( \frac{v^2}{r} = g \), and solve for \( v \), the linear velocity, using \( v = \sqrt{gr} \).
04

Calculate Orbital Velocity

Given \( g = 9.8 \, \text{m/s}^2 \) and \( r = 1.496 \times 10^{11} \, \text{m} \) (the average Earth-Sun distance), calculate the velocity: \( v = \sqrt{9.8 \times 1.496 \times 10^{11}} \approx 1.2 \times 10^4 \, \text{m/s} \).
05

Determine the Period of Revolution

Relate the linear velocity to the period of revolution \( T \) using \( v = \frac{2\pi r}{T} \). Solve for \( T \): \( T = \frac{2 \pi r}{v} \).
06

Calculate the Period in Days

Substitute \( r = 1.496 \times 10^{11} \, \text{m} \) and \( v \approx 1.2 \times 10^4 \, \text{m/s} \) into the formula for \( T \): \( T = \frac{2 \pi \times 1.496 \times 10^{11}}{1.2 \times 10^4} \approx 3.15 \times 10^7 \, \text{s} \). Convert seconds to days by dividing by \( 86400 \, \text{s/day} \), yielding approximately 365 days.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Artificial Gravity
In the world of science fiction and innovative engineering, artificial gravity represents a fascinating concept that allows us to imitate the gravitational pull found on Earth. This is particularly essential for long-term space travel or habitation on structures not possessing their own gravity. In the context of the artificial planet described in the scenario, artificial gravity is created by the rotation of a structure, such as the band encircling the sun.

The principle behind artificial gravity involves centripetal force. When the band rotates, the force pulling towards the center causes an acceleration that mimics gravity. This means the inhabitants of the band experience a force that feels like the familiar pull towards the ground. However, unlike Earth where gravity is due to mass, here it is the rotation itself creating the required centripetal acceleration.

To achieve this, the band must rotate with enough speed so that the acceleration equals the gravitational acceleration on Earth, denoted as \( g \). This ensures that the living conditions are comfortable and Earth-like, maintaining the necessary environment for any life forms present.
Orbital Mechanics
Orbital mechanics is the field of physics that explains how objects move in space under the influence of forces, primarily gravity. It is central to understanding scenarios like the artificial planet encircling a sun. In essence, orbital mechanics combines elements of physics and mathematics to predict and analyze the motion of celestial bodies.

For any object to maintain a stable orbit, like the band around the sun, it must follow Kepler’s laws of planetary motion and Newton’s law of gravitation. This means there needs to be a balance between the centripetal force, necessary to keep the object on its circular path, and the gravitational pull from the sun. The band’s rotation not only assists with artificial gravity, but also ensures it remains in orbit, maintaining a fixed distance - similar to how planets revolve around a star.

Understanding these concepts is pivotal for the design and execution of such projects, ensuring they remain stable and achieve the desired effect of both artificial gravity and secure orbit.
Period of Revolution
The period of revolution refers to the time it takes for an object to complete one full orbit around another object. In our specific scenario, we are interested in the period of revolution for the band around its sun, analogous to Earth's year length.

Calculating the period of revolution involves using the velocity of the band and the radius of its orbit. The relation is given by the formula \( T = \frac{2 \pi r}{v} \), where \( T \) is the period, \( r \) is the radius (distance from the sun), and \( v \) is the linear velocity of the band.

In our example, this computation is vital as it dictates how long a year lasts on this artificial planet. Since the sun in the scenario is like our own and the radius is similar to Earth's orbit, when the figures are plugged into the formula, the period of revolution conveniently results in approximately 365 Earth days. This demonstrates the balance maintained between synthetic gravitational fields and celestial mechanics, creating a year-length that supports a temperate climate conducive to life as we know it.

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Most popular questions from this chapter

The Navstar Global Positioning System (GPS) utilizes a group of 24 satellites orbiting the Earth. Using "triangulation" and signals transmitted by these satellites, the position of a receiver on the Earth can be determined to within an accuracy of a few centimeters. The satellite orbits are distributed evenly around the Earth, with four satellites in each of six orbits, allowing continuous navigational "fixes" The satellites orbit at an altitude of approximately \(11,000\) nautical miles \([1\) nautical mile \(=\) \(1.852 \mathrm{km}=6076 \mathrm{ft} \mathrm{t}\) . \((a)\) Determine the speed of each satellite. (b) Determine the period of each satellite.

A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the size of our Moon but has the mass of our Sun. ( \(a\) ) Estimate gravity on the surface on this star. \((b)\) How much would a \(65-\mathrm{kg}\) person weigh on this star? (c) What would be the speed of a baseball dropped from a height of \(1.0 \mathrm{~m}\) when it hit the surface?

(II) The accompanying table shows the data for the mean distances of planets (except Pluto) from the Sun in our solar system, and their periods of revolution about the Sun. $$ \begin{array}{lcc} \hline \text { Planet } & \text { Mean Distance (AU) } & \text { Period (Years) } \\ \hline \text { Mercury } & 0.387 & 0.241 \\ \text { Venus } & 0.723 & 0.615 \\ \text { Earth } & 1.000 & 1.000 \\ \text { Mars } & 1.524 & 1.881 \\ \text { Jupiter } & 5.203 & 11.88 \\ \text { Saturn } & 9.539 & 29.46 \\ \text { Uranus } & 19.18 & 84.01 \\ \text { Neptune } & 30.06 & 164.8 \\ \hline \end{array} $$ (a) Graph the square of the periods as a function of the cube of the average distances, and find the best-fit straight line. (b) If the period of Pluto is 247.7 years, estimate the mean distance of Pluto from the Sun from the best- fit line.

A satellite circles a spherical planet of unknown mass in a circular orbit of radius \(2.0 \times 10^{7} \mathrm{~m} .\) The magnitude of the gravitational force exerted on the satellite by the planet is 120 N. \((a)\) What would be the magnitude of the gravitational force exerted on the satellite by the planet if the radius of the orbit were increased to \(3.0 \times 10^{7} \mathrm{~m} ?(b)\) If the satellite circles the planet once every \(2.0 \mathrm{~h}\) in the larger orbit, what is the mass of the planet?

Estimate the value of the gravitational constant \(G\) in Newton's law of universal gravitation using the following data: the acceleration due to gravity at the Earth's surface is about \(10 \mathrm{~m} / \mathrm{s}^{2} ;\) the Earth has a circumference of about \(40 \times 10^{6} \mathrm{~m}\); rocks found on the Earth's surface typically have densities of about \(3000 \mathrm{~kg} / \mathrm{m}^{3}\) and assume this density is constant throughout (even though you suspect it is not true).

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